The Work Friction Takes

solved Work

Q. A body of mass 2 kg $m = 2$ kg initially at rest $v_{0} = 0$ m/s moves under the action of an applied horizontal force $F_{h} = | F_{h} |$ of 7 N $| F_{h} | = 7$ N on a table with coefficient of kinetic friction = 0.1 $F_{t} = −0.1 \cdot m \cdot g$ . Taking g as 9.8 m/s² $g = 9.8$ m/s², compute the work done by friction in 10 s $Δ t = 10$ s, in joules.

Inputs

mass

m

2 kg

force

| F_{h} |

7 N

gravity

g

9.8 m/s²

time

Δ t

10.00 s

Outputs

work

W_{F,t}

How it's solved

Solved with Work

Given

$m = 2$ kg
$| F_{h} | = 7$ N
$g = 9.8$ m/s²
$Δ t = 10$ s
$v_{0} = 0$ m/s from “initially at rest”
$F_{h} = | F_{h} |$ from “applied horizontal force”
$F_{t} = −0.1 \cdot m \cdot g$ from “coefficient of kinetic friction = 0.1”

Assumed

$x_{0} = 0$ m

Find $W_{F,t}$ .

Working

coefficient of kinetic friction = 0.1 $F_{t} = −0.1 \cdot m \cdot g$
$F_{t} = −0.1 \cdot 2 \cdot 9.8$
$F_{t} = −1.96$ N
applied horizontal force $F_{h} = | F_{h} |$
$F_{h} = 7$
$F_{h} = 7$ N
net force is the resultant(body, sliding, x) $F_{net} = F_{h} + F_{t} + 0$
$F_{net} = 7 + (−1.96) + 0$
$F_{net} = 5.04$ N
second law(sliding, x) $F_{net} = m \cdot a$
$5.04 = 2 \cdot a$
$a = 2.52$ m/s²
position after a time(sliding, x) $x_{1} = x_{0} + v_{0} \cdot Δ t + \frac{a \cdot {Δ t}^{2}}{2}$
$x_{1} = 0 + 0 \cdot 10 + \frac{2.52 \cdot 10^{2}}{2}$
$x_{1} = 126$ m
change of position(sliding, x) $Δ x = x_{1} - x_{0}$
$Δ x = 126 - 0$
$Δ x = 126$ m
work of a steady force(body, sliding, 2) $W_{F,t} = F_{t} \cdot Δ x$
$W_{F,t} = (−1.96) \cdot 126$
$W_{F,t} = −247$ J
$W_{F,t} \approx −247$ Jto 3 significant figures

Conclusion

Friction points against the motion, so its work is negative: it takes away energy the push puts in.

The web of concepts

given read from the words equation asked

The Ruby that solves it

The question and its solution are small Ruby files. The engine runs them.

question.rb

1require_relative "../physics"23question :the_work_friction_takes do4  kind :numerical5  describes "A body of mass %{mass} initially at rest moves under the action of an applied horizontal force of " \6            "%{force} on a table with coefficient of kinetic friction = 0.1. Taking g as %{gravity}, compute the " \7            "%{work} in %{time}, in joules."8  given mass: "2 kg"9  given force: "7 N"10  given gravity: "9.8 m/s2"11  given time: "10 s"12  asks work: "work done by friction"13end

solution.rb

1require_relative "../questions/the_work_friction_takes"2require_relative "../science/work_energy_and_power/work"34solve :the_work_friction_takes, using: Work do5  objects :body, :hand, :table6  axes :x7  body.goes(:sliding)89  take mass: body.mass10  take force: body.force(by: hand).magnitude11  take gravity: g12  take time: sliding.time_taken1314  assume body.initial_position => "0 m"15  deduce_from "initially at rest", body.initial_velocity => "0 m/s"16  deduce_from "applied horizontal force", body.force(by: hand).direction => x17  deduce_from "coefficient of kinetic friction = 0.1", body.force(by: table) => -0.1 * body.mass * g1819  work = body.work(by: table)20  answer work:, in: "J", figures: 32122  conclusion "Friction points against the motion, so its work is negative: it takes away energy the push puts in."23end

playground.rb

1require_relative "../solutions/the_work_friction_takes"23play_with :the_work_friction_takes do4  title "The Work Friction Takes"5end

The laws it drew on

quantity	written	units	within	from
volume	$volume$	—	—	Vectors
mass	$m$	kg	0 … ∞	Vectors
speed	$speed$	—	—	Vectors
density	$density$	—	—	Vectors
number of moles	$n$	—	—	Vectors
angular frequency	${angular}_{frequency}$	—	—	Vectors
work	$W_{F}$	J	—	Vectors
current	$current$	—	—	Vectors
temperature	$temperature$	—	—	Vectors
pressure	$pressure$	—	—	Vectors
time	$t$	s	0 … 1e+05	Vectors
power	$power$	—	—	Vectors
total path length	$s$	—	—	Vectors
energy	$energy$	—	—	Vectors
gravitational potential	${gravitational}_{potential}$	—	—	Vectors
coefficient of friction	$μ$	—	—	Vectors
charge	$charge$	—	—	Vectors
acceleration	$a$	m/s²	—	Vectors
velocity	$v$	m/s	—	Vectors
displacement	$Δ x$	m	—	Vectors
angular velocity	${angular}_{velocity}$	—	—	Vectors
force	$F$	N	—	Vectors
angular momentum	${angular}_{momentum}$	—	—	Vectors
linear momentum	$p_{lin}$	—	—	Vectors
electric field	${electric}_{field}$	—	—	Vectors
average velocity	$v_{av}$	m/s	—	Vectors
magnetic moment	${magnetic}_{moment}$	—	—	Vectors
relative velocity	$v_{AB}$	—	—	Vectors
impulse	$impulse$	—	—	Vectors
first vector	$A$	—	0 … ∞	Vectors
second vector	$B$	—	0 … ∞	Vectors
angle between	$φ$	—	0 … 3.142	Vectors
resultant	$R$	—	0 … ∞	Vectors
resultant direction	$β$	—	0 … 3.142	Vectors
direction	$α$	—	-3.142 … 3.142	Vectors
x component	$A_{x}$	—	—	Vectors
y component	$A_{y}$	—	—	Vectors
position	$x$	m	—	Velocity
path length	$d$	m	—	Velocity
time taken	$Δ t$	s	0 … 1e+05	Velocity
average speed	$c_{av}$	m/s	—	Velocity
starting position	$x_{0}$	m	—	Velocity
final position	$x$	m	—	Velocity
change of velocity	$Δ v$	m/s	—	Acceleration
average acceleration	$a_{av}$	m/s²	—	Acceleration
starting velocity	$v_{0}$	m/s	—	Acceleration
final velocity	$v$	m/s	—	Acceleration
gravity	$g$	m/s²	—	UniformAcceleration
net force	$F_{net}$	N	—	NewtonsSecondLaw
weight	$W$	N	—	NewtonsSecondLaw
work by gravity	$W_{g}$	J	—	Work
net work	$W_{net}$	J	—	Work
displacement of body during sliding along x	$Δ x$	m	—
position of body at end of sliding along x	$x_{1}$	m	—
position of body at start of sliding along x	$x_{0}$	m	—
average velocity of body during sliding along x	$v_{av}$	m/s	—
time taken of sliding	$Δ t$	s	0 … 1e+05
average speed of body during sliding	$c_{av}$	m/s	—
path length of body during sliding	$d$	m	—
change of velocity of body during sliding along x	$Δ v$	m/s	—
velocity of body at end of sliding from ground along x	$v_{1}$	m/s	—
velocity of body at start of sliding from ground along x	$v_{0}$	m/s	—
average acceleration of body during sliding along x	$a_{av}$	m/s²	—
acceleration of body during sliding along x	$a$	m/s²	—
net force on body during sliding along x	$F_{net}$	N	—
mass of body	$m$	kg	0 … ∞
mass of hand	$m_{h}$	kg	0 … ∞
mass of table	$m_{t}$	kg	0 … ∞
weight on ground	$\| W_{g} \|$	N	—
weight on body	$\| W \|$	N	—
weight on hand	$\| W_{h} \|$	N	—
weight on table	$\| W_{t} \|$	N	—
force on body by hand during sliding	$\| F_{h} \|$	N	—
force on body by table during sliding	$\| F_{t} \|$	N	—
force on body by hand during sliding along x	$F_{h}$	N	—
force on body by table during sliding along x	$F_{t}$	N	—
work on body by hand during sliding	$W_{F,h}$	J	—
work on body by table during sliding	$W_{F,t}$	J	—
net work on body during sliding	$W_{net}$	J	—
displacement of body during sliding	$\| Δ x \|$	m	—
position of body at end of sliding	$\| x_{1} \|$	m	—
position of body at start of sliding	$\| x_{0} \|$	m	—
average velocity of body during sliding	$\| v_{av} \|$	m/s	—
change of velocity of body during sliding	$\| Δ v \|$	m/s	—
velocity of body at end of sliding from ground	$\| v_{1} \|$	m/s	—
velocity of body at start of sliding from ground	$\| v_{0} \|$	m/s	—
average acceleration of body during sliding	$\| a_{av} \|$	m/s²	—
acceleration of body during sliding	$\| a \|$	m/s²	—
net force on body during sliding	$\| F_{net} \|$	N	—

x_{0} \equiv x_{i} (start of motion)

starting position

x \equiv x_{i} (end of motion)

final position

v_{0} \equiv v_{i} (start of motion)

starting velocity

v \equiv v_{i} (end of motion)

final velocity

t \equiv Δ t_{m}

time

A = \sqrt{{A_{x}}^{2} + {A_{y}}^{2}}

magnitude from components

d_{i} (whole) = d_{i} (part 1) + d_{i} (part 2)

path in parts

Δ t_{w} = Δ t_{p} + Δ t_{p}

time in parts

a_{i} (part 1) = a_{i} (whole)

steady through its parts

m_{b} \neq 0

has mass?

R^{2} = A^{2} + B^{2} + 2 \cdot A \cdot B \cdot \cos φ

law of cosines

R \cdot \sin β = B \cdot \sin φ

law of sines

R \cdot \cos β = A + B \cdot \cos φ

along the first

A_{x} = A \cdot \cos α

resolved along x

A_{y} = A \cdot \sin α

resolved along y

Δ x = x_{1} - x_{0}

change of position(sliding, x)

v_{av} = \frac{Δ x}{Δ t}

displacement a time(sliding, x)

c_{av} = \frac{d}{Δ t}

path a time(sliding, x)

Δ v = v_{1} - v_{0}

velocity gained(sliding, x)

a_{av} = \frac{Δ v}{Δ t}

velocity gained a time(sliding, x)

v_{1} = v_{0} + a \cdot Δ t

velocity after a time(sliding, x)

x_{1} = x_{0} + v_{0} \cdot Δ t + \frac{a \cdot {Δ t}^{2}}{2}

position after a time(sliding, x)

{v_{1}}^{2} = {v_{0}}^{2} + 2 \cdot a \cdot (x_{1} - x_{0})

velocity after a distance(sliding, x)

x_{1} - x_{0} = \frac{(v_{1} + v_{0}) \cdot Δ t}{2}

mean of the velocities(sliding, x)

F_{net} = m \cdot a

second law(sliding, x)

F_{net} = F_{h} + F_{t} + 0

net force is the resultant(body, sliding, x)

W_{F,h} = F_{h} \cdot Δ x

work of a steady force(body, sliding, 1)

W_{F,t} = F_{t} \cdot Δ x

work of a steady force(body, sliding, 2)

W_{net} = F_{net} \cdot Δ x

work of the net force(body, sliding)

| Δ x | = \sqrt{{Δ x}^{2}}

magnitude from components(displacement of body during sliding)

| x_{1} | = \sqrt{{x_{1}}^{2}}

magnitude from components(position of body at end of sliding)

| x_{0} | = \sqrt{{x_{0}}^{2}}

magnitude from components(position of body at start of sliding)

| v_{av} | = \sqrt{{v_{av}}^{2}}

magnitude from components(average velocity of body during sliding)

| Δ v | = \sqrt{{Δ v}^{2}}

magnitude from components(change of velocity of body during sliding)

| v_{1} | = \sqrt{{v_{1}}^{2}}

magnitude from components(velocity of body at end of sliding from ground)

| v_{0} | = \sqrt{{v_{0}}^{2}}

magnitude from components(velocity of body at start of sliding from ground)

| a_{av} | = \sqrt{{a_{av}}^{2}}

magnitude from components(average acceleration of body during sliding)

| a | = \sqrt{a^{2}}

magnitude from components(acceleration of body during sliding)

| F_{net} | = \sqrt{{F_{net}}^{2}}

magnitude from components(net force on body during sliding)

| F_{h} | = \sqrt{{F_{h}}^{2}}

magnitude from components(force on body by hand during sliding)

| F_{t} | = \sqrt{{F_{t}}^{2}}

magnitude from components(force on body by table during sliding)

F_{h} = | F_{h} |

applied horizontal force

F_{t} = −0.1 \cdot m \cdot g

coefficient of kinetic friction = 0.1

m \neq 0

has mass?(body)

m_{h} \neq 0

has mass?(hand)

m_{t} \neq 0

has mass?(table)

F_{net} = 0

free of forces?(body, sliding)